经典sql面试题

有3个表S，C，SC

S（SNO，SNAME）代表（学号，姓名）

C（CNO，CNAME，CTEACHER）代表（课号，课名，教师）

SC（SNO，CNO，SCGRADE）代表（学号，课号，成绩）

问题：

1，找出没选过“黎明”老师的所有学生姓名。

2，列出2门以上（含2门）不及格学生姓名及平均成绩。

3，既学过1号课程又学过2号课所有学生的姓名。

CREATE TABLE SC
(
SNO VARCHAR(200),
CNO VARCHAR(200),
SCGRADE VARCHAR(200)
);
CREATE TABLE S
(
SNO VARCHAR(200),
SNAME VARCHAR(200)
);
CREATE TABLE C
(
CNO VARCHAR(200),
CNAME VARCHAR(200),
CTEACHER VARCHAR(200)
);

INSERT INTO C ( CNO, CNAME, CTEACHER ) VALUES ( '1', '语文', '张');
INSERT INTO C ( CNO, CNAME, CTEACHER ) VALUES ( '2', '政治', '王');
INSERT INTO C ( CNO, CNAME, CTEACHER ) VALUES ( '3', '英语', '李');
INSERT INTO C ( CNO, CNAME, CTEACHER ) VALUES ( '4', '数学', '赵');
INSERT INTO C ( CNO, CNAME, CTEACHER ) VALUES ( '5', '物理', '黎明');
commit;

INSERT INTO S ( SNO, SNAME ) VALUES ( '1', '学生1');
INSERT INTO S ( SNO, SNAME ) VALUES ( '2', '学生2');
INSERT INTO S ( SNO, SNAME ) VALUES ( '3', '学生3');
INSERT INTO S ( SNO, SNAME ) VALUES ( '4', '学生4');
commit;

INSERT INTO SC ( SNO, CNO, SCGRADE ) VALUES ( '1', '1', '40');
INSERT INTO SC ( SNO, CNO, SCGRADE ) VALUES ( '1', '2', '30');
INSERT INTO SC ( SNO, CNO, SCGRADE ) VALUES ( '1', '3', '20');
INSERT INTO SC ( SNO, CNO, SCGRADE ) VALUES ( '1', '4', '80');
INSERT INTO SC ( SNO, CNO, SCGRADE ) VALUES ( '1', '5', '60');
INSERT INTO SC ( SNO, CNO, SCGRADE ) VALUES ( '2', '1', '60');
INSERT INTO SC ( SNO, CNO, SCGRADE ) VALUES ( '2', '2', '60');
INSERT INTO SC ( SNO, CNO, SCGRADE ) VALUES ( '2', '3', '60');
INSERT INTO SC ( SNO, CNO, SCGRADE ) VALUES ( '2', '4', '60');
INSERT INTO SC ( SNO, CNO, SCGRADE ) VALUES ( '2', '5', '40');
INSERT INTO SC ( SNO, CNO, SCGRADE ) VALUES ( '3', '1', '60');
INSERT INTO SC ( SNO, CNO, SCGRADE ) VALUES ( '3', '3', '80');
commit;1、求黎明老师教的所有课的课号

select sname from s where sno not in
(
select sno from sc where cno in
(
select distinct cno from c where cteacher='黎明'
)
);2、列出2门以上（含2门）不及格学生姓名及平均成绩。

Select sname from s where sno in (select sno from sc where scgrade<60 group by sno having count(*)>=2)

select s.sname, avg_grade from s
join
(select sno from sc where scgrade < 60 group by sno having count(*) >= 2) t1
on s.sno = t1.sno
join
(select sno, avg(scgrade) as avg_grade from sc group by sno ) t2
on s.sno = t2.sno;

3、既学过1号课程又学过2号课所有学生的姓名

select sname from s where
sno in (select sno from sc where cno = 1)
and
sno in (select sno from sc where cno = 2);